Question: $\dfrac{d}{dx}[\cos(5x-9)]=\,?$ Choose 1 answer: Choose 1 answer: (Choice A) A $-5\sin(x)$ (Choice B) B $-5\sin(5x-9)$ (Choice C) C $-(5x-9)\sin(5)$ (Choice D) D $-\sin(5x-9)$
Since $\cos(5x-9)$ is a composite function, we can use the chain rule. The chain rule says: $\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right]={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)}$ To use it, we need to recognize our function as a composite function like $w\big(u(x)\big)$. $\underbrace{\cos(~\overbrace{5x-9}^{\text{inner}}~)}_{\text{outer}}$ So if $\cos(5x-9)=w(u(x))$, then: $\begin{aligned} {u(x)}&={5x-9} &&\text{inner function} \\\\ w(x)&=\cos(x)&&\text{outer function} \end{aligned}$ Before applying the chain rule, let's find the derivatives of the inner and outer functions (work not shown, but hopefully these derivatives are familiar by now). $\begin{aligned} {u'(x)}&={5} \\\\ {w'(x)}&={-\sin(x)} \end{aligned}$ Now let's apply the chain rule: $\begin{aligned} \dfrac{d}{dx}[\cos(5x-9)]&=\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right] \\\\ &={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)} \\\\ &={-\sin({5x-9})} \cdot {5} \\\\ &=-5\sin(5x-9) \end{aligned}$